JEE Physics Mechanical Properties Of Fluids MCQs Set 26

Answer: (a) \( 8.0\times10^{-3} \) m³/s

Explanation:
1. Area at the wide section = \( \pi(0.025)^2=1.96\times10^{-3} \) m².
2. Q = Av = \( 1.96\times10^{-3}\times4.08\approx8.0\times10^{-3} \) m³/s.

Teacher's Note:
1. Use the radius, not the diameter, in the area.
2. The same flow rate passes through the throat at five times the speed.

Answer: (b) 3.75 kg

Explanation:
1. Pressure is the same: \( m_1g/A_1=m_2g/A_2 \).
2. Area ratio = 400 : 1, so \( m_1=1500/400=3.75 \) kg.

Teacher's Note:
1. Area scales as the square of the radius.
2. Force is multiplied by the area ratio.

Answer: (b) Only (1) and (3)

Explanation:
1. Pressure decreases with height: \( \partial p/\partial z=-dg \).
2. There is no variation along the horizontal at a given level: \( \partial p/\partial x=0 \).

Teacher's Note:
1. The pressure gradient in a static fluid is vertical only.
2. The negative sign shows that pressure falls as z increases.

Answer: (c) \( (v-V)^2 \)

Explanation:
1. Relative speed of the jet = v − V.
2. Mass arriving per second = \( \rho A(v-V) \) and each unit loses momentum \( (v-V) \).
3. F = \( \rho A(v-V)^2 \).

Teacher's Note:
1. The rate of mass arrival also depends on the relative speed.
2. If the plate is at rest (V = 0) the force is \( \rho Av^2 \).

Answer: (a) \( \dfrac{5d}{4} \)

Explanation:
1. Weight = buoyancy: \( D\,L=\tfrac{3L}{4}(d)+\tfrac{L}{4}(2d) \).
2. \( D=\tfrac34d+\tfrac12d=\tfrac54d \).

Teacher's Note:
1. Each part of the cylinder displaces the liquid in which it lies.
2. The density lies between d and 2d, as it must to float at the interface.

Answer: (c) 6 × 1000 × 10 × 80 N

Explanation:
1. Pressure at the bottom = \( \rho g h=1000\times10\times6 \).
2. Thrust = pressure × area = \( 6\times1000\times10\times(10\times8) \) N.

Teacher's Note:
1. On a horizontal base the thrust is the full depth pressure times the whole area.
2. The factor 3 would apply to a side wall (average pressure).

Answer: (a) 6 × 5 × 5 × 5 gf

Explanation:
1. Weight in air = 125 × 7 gf.
2. Buoyancy = 125 × 1 gf.
3. Apparent weight = 125(7 − 1) = 6 × 5 × 5 × 5 gf.

Teacher's Note:
1. Apparent weight = (ρ − ρw)Vg.
2. In CGS, 1 cc of water weighs 1 gf.

Answer: (d) 15 m

Explanation:
1. Thrust on curved surface = (average pressure) × area = \( \rho g(10+5)\times2\pi R(10) \).
2. Thrust on the bottom = \( \rho g(10+10)\times\pi R^2 \).
3. Equating: \( 300R=20R^2 \), so R = 15 m.

Teacher's Note:
1. Atmospheric pressure acts on the outer and inner faces and must be included, as the thrust here is the total.
2. Average pressure on the curved wall is taken at mid-depth.

Answer: (b) 500 kg/m³ and 0.5

Explanation:
1. Gauge pressure \( \rho g h=15000 \) N/m².
2. \( \rho=\dfrac{15000}{10\times3}=500 \) kg/m³.
3. Specific gravity = 500/1000 = 0.5.

Teacher's Note:
1. Gauge pressure is the excess over atmospheric pressure, so no atmospheric term is needed.
2. Specific gravity is density divided by the density of water.

Answer: (d) 6 mm

Explanation:
1. In the steady state the trapped air pressure is \( P=P_0-\rho gh=10^5-1000(10)(0.2)=98000 \) Pa.
2. Boyle's law: \( P_0(500-H)=98000(300) \), so \( 500-H=294 \) mm.
3. H = 206 mm. The fall in level = 206 − 200 = 6 mm.

Teacher's Note:
1. The air above the water expands until its pressure plus the water column's pressure equals atmospheric pressure.
2. Use lengths in mm since only ratios matter.

Answer: (b) 1.48 g/cm³

Explanation:
1. Weight = buoyancy: \( \rho\cdot\tfrac43\pi(10^3-9^3)g=0.8\cdot\tfrac23\pi(10^3)g \).
2. \( \rho\cdot\tfrac43(271)=0.8\cdot\tfrac23(1000) \), so \( \rho=\dfrac{800}{542}=1.476 \) g/cm³.

Teacher's Note:
1. The volume of material is the difference of the two sphere volumes.
2. The displaced volume is half of the outer sphere.

Answer: (d) 0.25 m/s, 0.50 m/s, 93.75 Pa

Explanation:
1. \( v_x=10\times10^{-6}/40\times10^{-6}=0.25 \) m/s and \( v_y=0.50 \) m/s.
2. \( P_x-P_y=\tfrac12\rho(v_y^2-v_x^2)=\tfrac12(1000)(0.25-0.0625)=93.75 \) Pa.

Teacher's Note:
1. Convert cm³ to m³ and mm² to m² carefully.
2. Pressure is higher where the tube is wider.

Answer: (d) \( L/\sqrt{2\pi} \)

Explanation:
1. Flow rate = area × \( \sqrt{2g\cdot\text{depth}} \).
2. \( L^2\sqrt{y}=\pi R^2\sqrt{4y}=2\pi R^2\sqrt y \).
3. \( R^2=\dfrac{L^2}{2\pi} \), so \( R=\dfrac{L}{\sqrt{2\pi}} \).

Teacher's Note:
1. Efflux speed varies as the square root of the depth, so the deeper hole is twice as fast.
2. Equate area times speed for the two holes.

Answer: (b) h

Explanation:
1. Thrust from a hole = \( \rho a v^2=2\rho a g\,(\text{depth}) \).
2. Net force = \( 2\rho a g(d_1-d_2)=2\rho a g h \), proportional to h.

Teacher's Note:
1. The two jets give opposite thrusts, so only the depth difference remains.
2. Thrust is proportional to depth because it varies as v².

Answer: (b) 57 cm

Explanation:
1. \( (D/D_0)^4=b/(b+x) \): \( 0.9^4=0.6561 \).
2. \( b=\dfrac{0.6561\,x}{1-0.6561}=\dfrac{0.6561(0.3)}{0.3439}=0.572 \) m ≈ 57 cm.

Teacher's Note:
1. Raise the diameter ratio to the fourth power.
2. Convert 0.572 m to cm.

Answer: (d) 40 g more than R

Explanation:
1. The ball experiences a buoyant force of 40 gf (weight of 40 cm³ of water).
2. By Newton's third law the water is pushed down with 40 gf.
3. So the beaker pan needs 40 g more on the other side.

Teacher's Note:
1. The buoyant force on the ball has an equal and opposite reaction on the liquid.
2. The weight of the ball is carried by the thread, not by the pan.

Answer: (c) 1.0 m/s

Explanation:
1. Continuity: \( A(3)=A(1.5)+1.5A\,v \).
2. \( 1.5=1.5v \), so v = 1.0 m/s.

Teacher's Note:
1. Volume flow rate entering equals the sum leaving.
2. Each branch contributes area times speed.

Answer: (c) zero

Explanation:
1. \( P_x+\tfrac12\rho v_x^2+\rho gh=P_y+\tfrac12\rho v_y^2 \).
2. \( P_x-P_y=93.75-1000(10)(0.009375)=93.75-93.75=0 \).

Teacher's Note:
1. The kinetic energy gain is exactly paid by the loss in height energy.
2. Both terms equal 93.75 Pa here.

Answer: (d) \( (H-h)\rho g \)

Explanation:
1. The depth of P below the surface is H − h.
2. Pressure (thrust per unit area, acting normal to the wall) = \( (H-h)\rho g \).

Teacher's Note:
1. Pressure depends only on depth and is independent of the wall inclination.
2. Use depth from the free surface, not height from the base.

Answer: (c) \( 6.4\times10^{-4} \) m³/s

Explanation:
1. Continuity: \( v_2=v_1(A_1/A_2)=6.25\,v_1 \).
2. Bernoulli: \( \Delta P=\tfrac12\rho(v_2^2-v_1^2)=\tfrac12\rho v_1^2(6.25^2-1) \), giving \( v_1=0.0205 \) m/s.
3. Q = \( A_1v_1=\pi(0.1)^2(0.0205)\approx6.4\times10^{-4} \) m³/s.

Teacher's Note:
1. The wide end has the larger pressure and the smaller speed.
2. Always convert the pressure drop through Bernoulli before using continuity.

Answer: (c) \( t_1\,\dfrac{d_L}{d_L-d} \)

Explanation:
1. Time in air (fall and return) = \( t_1 \); speed on reaching the liquid is \( v=gt_1/2 \).
2. In the liquid the retardation is \( a=g(d_L-d)/d \), so the time in the liquid is \( 2v/a=\dfrac{t_1d}{d_L-d} \).
3. \( t_2=t_1+\dfrac{t_1d}{d_L-d}=\dfrac{t_1d_L}{d_L-d} \).

Teacher's Note:
1. The net upward force in the liquid is \( Vg(d_L-d) \) and the mass is \( Vd \).
2. In the liquid the ball moves like a particle with constant retardation.

Answer: (c) 3

Explanation:
1. Buoyant force = 60 − 40 = 20 N.
2. \( \dfrac{\text{weight}}{\text{buoyancy}}=\dfrac{\rho_b}{\rho_w}=\dfrac{60}{20}=3 \).

Teacher's Note:
1. Weight over loss of weight gives the relative density.
2. The submerged volume equals the block's volume.

Answer: (a) Shyam has more fat content than Ram

Explanation:
1. Under-water reading = weight − buoyancy.
2. A larger reading means less buoyancy, so less volume and a higher average density for Ram.
3. So Ram has less fat; Shyam has more.

Teacher's Note:
1. Fat lowers the average density and increases the volume for a given mass.
2. A larger volume gives more buoyancy.

Answer: (a) F/A

Explanation:
1. Pressure change applied to an enclosed liquid is transmitted undiminished (Pascal's law).
2. Each piston raises the pressure by F/A and the equilibrium of every piston requires the same rise, so the increase is F/A.

Teacher's Note:
1. The pressure increases are not added piston by piston, because equilibrium of the pistons fixes a single value.
2. Pressure is a scalar and has no direction.

Answer: (a) 10 cm

Explanation:
1. Extra buoyant force that supported the 200 g mass = weight of the extra water displaced: \( 200=1\times a^2\times2 \) (in g, cm).
2. \( a^2=100 \), so a = 10 cm.

Teacher's Note:
1. The extra immersed volume equals the load divided by the density of water.
2. Work in g and cm to avoid conversions.